No. Pointers are pointers whatever they point at, be it an array or a structure. Think about a piece of memory as a sequence of cells. So a pointer to any cell allows you to access all the following cells in order, by using an index (pointer plus offset). If you (the programmer) have decided that the area you point to should be treated as if it contains different sized blocks (aka a structure), that does not affect the physical properties of the memory. It merely allows the compiler to calculate the distance between the elements of the structure. And an array of pointers is much the same thing.

If you have trouble visualising multi levels of indirection, then always go for a single level. If you have an array of pointers, then create a temporary one and allocate an array entry to it like:

CUSTOMVERTEX ** ScreenLettersP_s = new CUSTOMVERTEX* [NumberOfTextBuffers]; // an array of struct pointers
CUSTOMVERTEX* pTemp = ScreenLettersP_s[0]; // get the first pointer in the array
pTemp-> // now access the struct items.

I have not used DirectX, but I have used plenty of other Windows' functions that use structures, arrays of structures, and even arrays of structures that contain other unstructured structures. In the latter case, the presence or absence of certain items depends on settings elsewhere.

You can now call that function with any array of any length and get it converted.
In every case the function receives the physical address of the array and accesses each
character by using the index value i, where 0 <= i < length.

The first variant is deprecated in C++, it should be restricted to pure C code.
The second variant is useful if you know the size of your arrays at compile time (and it's always the same)
The third variant is the most flexible as you don't need to know the array size, and you can even add more values within your function if you desire.

GOTOs are a bit like wire coat hangers: they tend to breed in the darkness, such that where there once were few, eventually there are many, and the program's architecture collapses beneath them. (Fran Poretto)

There are a total of P levels. For any level p, there are Jp nodes. Any two nodes on the same level are not connected. And the unidirectional path weight between any two nodes Ai and Bj is D(i,j).
What is the shortest distance required from the first layer to the Pth layer?
(** Note that P is generally large and it is not recommended to traverse all possible situations **)
If this problem is abstracted as a graph, then what is actually required is the shortest path from the first layer to the last layer of a fully connected network.

What I have tried:
Use deep search, Dijkstra algorithm, Floydw algorithm, etc. to search the shortest path from all J1 nodes in the first layer to the last layer in turn, and select the minimum value from the J1 values.
What I want to know is if there is a simpler algorithm to solve this problem?

If I understand this correctly, you're looking for a unidirectional path from the first layer to the pth layer, a path that includes each layer only once.

I would start with the second layer. Assign each node its shortest distance from the first layer, and record the first layer node whose edge was selected. Then proceed through each layer in the same way: for each node, find its shortest distance from the nodes in the previous layer (the current weight of a previous node plus its edge to the current layer). Eventually you will find the shortest distances to nodes in the pth layer, and thus the shortest path to that layer.

Unfortunately it can in general not guarantee that it will find the global optimum. You can find a good guess very quickly, or you can slow the annealing to the pace of continental drifts and have a higher chance to find the correct answer, but you can never be 100% sure unless you test all options, i. e. brute force.

GOTOs are a bit like wire coat hangers: they tend to breed in the darkness, such that where there once were few, eventually there are many, and the program's architecture collapses beneath them. (Fran Poretto)